想到了用哈希表存这个数的位置,但是没有想到可以直接用哈希表存next great,用栈存还没找到的数,没遍历一个数就考察栈中的元素小,小的话,这个数就是栈中数的next great,栈中的数肯定是下大上小。 publicint[] nextGreaterElement(int[] nums1,int[] nums2) {/*通过map建立当前元素和其next great的映射 ...
Next Greater Element II Given a circular array (the next element of the last element is the first element of the array), print the Next Greater Number for every element. The Next Greater Number of a number x is the first greater number to its traversing-order next in the array, which m...
LeetCode题目:503. Next Greater Element II Given acircular array (the next element of the last element is the first element of the array),print the Next Greater Number for every element. The Next Greater Number of a number x is the first greater number to its traversing-order next in the...
503. Next Greater Element II # 题目 # Given a circular array (the next element of the last element is the first element of the array), print the Next Greater Number for every element. The Next Greater Number of a number x is the first greater number to
code: ``` class Solution(object): def nextGreaterElement(self, findNums, nums): """ :type findNums: List[int] :type nums: List[int] :rtype: List[int] """ dic = {} s = [] i = 0 while i != len(nums): if not s or s[-1] > nums[i]: ...
题目地址:https://leetcode.com/problems/minesweeper/description/ 题目描述 Given a circular array (the next element of the last element is the first element of the array), print the Next Greater Number for every element. The Next Greater Number of a number x is the first grea...
LeetCode-Next Greater Element I Description: You are given two arrays (without duplicates) nums1 and nums2 where nums1’s elements are subset of nums2. Find all the next greater numbers for nums1’s elements in the corresponding places of nums2....
简介:LeetCode之Next Greater Element I 1、题目 You are given two arrays (without duplicates) nums1 and nums2 where nums1’s elements are subset of nums2. Find all the next greater numbers for nums1's elements in the corresponding places of nums2.The Next Greater Number of a number x in...
1. 环状数组的一个经典处理方法是, 复制一份相同的追加到后面 2. 找一侧恰好比当前数大可以用单调栈 */classSolution {publicint[] nextGreaterElements(int[] nums) {if(nums == null || nums.length ==0)returnnums;intn = nums.length;int[] range =newint[n *2];for(inti =0; i < n; i+...
若栈非空,则此时栈顶数即为cur对应位置后最近较大数 将以上结果存入字典hashTable中,便于生成结果 nums2遍历结束后,遍历nums1 以当前遍历数,向hashTable中取值,并存入结果列表res 最后返回res 代码如下: class Solution(object): def nextGreaterElement(self,nums1,nums2): ...